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Questions: Algebra BusinessCalculus
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Find the equation of the line perpendicular to \(\displaystyle y = - \frac{x}{2} + 2\) that contains \(\displaystyle \left( 5, \ -4\right)\).
The slope of the line is \(\displaystyle m = - \frac{1}{2}\) so the perpendicular line has slope \(\displaystyle m=2\). The equation has the form \(\displaystyle y=2x+b\). Using the point \(\displaystyle \left( 5, \ -4\right)\) gives the equation \(\displaystyle -4=2\left(5\right)+b\) Solving for \(\displaystyle b\) gives \(\displaystyle b = -14\). The equation of the perpendicular line is \(\displaystyle y = 2 x - 14\).
\begin{question}Find the equation of the line perpendicular to $y = - \frac{x}{2} + 2$ that contains $\left( 5, \ -4\right)$.
\soln{9cm}{The slope of the line is $m = - \frac{1}{2}$ so the perpendicular line has slope $m=2$. The equation has the form $y=2x+b$. Using the point $\left( 5, \ -4\right)$ gives the equation $-4=2\left(5\right)+b$ Solving for $b$ gives $b = -14$. The equation of the perpendicular line is $y = 2 x - 14$. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Find the equation of the line perpendicular to <img class="equation_image" title=" \displaystyle y = - \frac{x}{2} + 2 " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20-%20%5Cfrac%7Bx%7D%7B2%7D%20%2B%202%20" alt="LaTeX: \displaystyle y = - \frac{x}{2} + 2 " data-equation-content=" \displaystyle y = - \frac{x}{2} + 2 " /> that contains <img class="equation_image" title=" \displaystyle \left( 5, \ -4\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%205%2C%20%5C%20%20-4%5Cright%29%20" alt="LaTeX: \displaystyle \left( 5, \ -4\right) " data-equation-content=" \displaystyle \left( 5, \ -4\right) " /> . </p> </p><p> <p>The slope of the line is <img class="equation_image" title=" \displaystyle m = - \frac{1}{2} " src="/equation_images/%20%5Cdisplaystyle%20m%20%3D%20-%20%5Cfrac%7B1%7D%7B2%7D%20" alt="LaTeX: \displaystyle m = - \frac{1}{2} " data-equation-content=" \displaystyle m = - \frac{1}{2} " /> so the perpendicular line has slope <img class="equation_image" title=" \displaystyle m=2 " src="/equation_images/%20%5Cdisplaystyle%20m%3D2%20" alt="LaTeX: \displaystyle m=2 " data-equation-content=" \displaystyle m=2 " /> . The equation has the form <img class="equation_image" title=" \displaystyle y=2x+b " src="/equation_images/%20%5Cdisplaystyle%20y%3D2x%2Bb%20" alt="LaTeX: \displaystyle y=2x+b " data-equation-content=" \displaystyle y=2x+b " /> . Using the point <img class="equation_image" title=" \displaystyle \left( 5, \ -4\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%205%2C%20%5C%20%20-4%5Cright%29%20" alt="LaTeX: \displaystyle \left( 5, \ -4\right) " data-equation-content=" \displaystyle \left( 5, \ -4\right) " /> gives the equation <img class="equation_image" title=" \displaystyle -4=2\left(5\right)+b " src="/equation_images/%20%5Cdisplaystyle%20-4%3D2%5Cleft%285%5Cright%29%2Bb%20" alt="LaTeX: \displaystyle -4=2\left(5\right)+b " data-equation-content=" \displaystyle -4=2\left(5\right)+b " /> Solving for <img class="equation_image" title=" \displaystyle b " src="/equation_images/%20%5Cdisplaystyle%20b%20" alt="LaTeX: \displaystyle b " data-equation-content=" \displaystyle b " /> gives <img class="equation_image" title=" \displaystyle b = -14 " src="/equation_images/%20%5Cdisplaystyle%20b%20%3D%20-14%20" alt="LaTeX: \displaystyle b = -14 " data-equation-content=" \displaystyle b = -14 " /> . The equation of the perpendicular line is <img class="equation_image" title=" \displaystyle y = 2 x - 14 " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%202%20x%20-%2014%20" alt="LaTeX: \displaystyle y = 2 x - 14 " data-equation-content=" \displaystyle y = 2 x - 14 " /> . </p> </p>